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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{1}{{{{({e^x} + {e^{ - x}})}^2}}}\;dx = } $
  • ✓
    $ - \frac{1}{{2({e^{2x}} + 1)}} + c$
  • B
    $\frac{1}{{2({e^{2x}} + 1)}} + c$
  • C
    $ - \frac{1}{{{e^{2x}} + 1}}$
  • D
    None of these
✓

Answer

Correct option: A.
$ - \frac{1}{{2({e^{2x}} + 1)}} + c$
a
(a)$\int_{}^{} {\frac{1}{{{{({e^x} + {e^{ - x}})}^2}}}\,dx = \int_{}^{} {\frac{{{e^{2x}}}}{{{{({e^{2x}} + 1)}^2}}}} \,dx} $
Put ${e^{2x}} + 1 = t \Rightarrow 2{e^{2x}}dx = dt,$ then it reduces to
$\frac{1}{2}\int_{}^{} {\frac{1}{{{t^2}}}dt} = - \frac{1}{2}.\frac{1}{t} + c = - \frac{1}{{2({e^{2x}} + 1)}} + c.$

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